trace , determinant and which of the following are true(NBHM-$2014$)

Question

Let $A \in M_2(\mathbb R)$ be a matrix which is not a diagonal matrix . Which of the following statements are true??

a. If $tr(A)=-1$ and $detA=1$, then $A^3=I$.

b. If $A^3=I$, then $tr(A)=-1$ and $det(A)=1$.

c. If $A^3=I$, then $A$ is diagonalizable over $\mathbb R$.

For (a), it is clear that $A$ will satisfy $\lambda^2+\lambda+1=0$ giving $A^2+A+I=0$. Multiplying $A$ through out gives $A^3+A^2+A=0\implies A^3=-A^2-A=I$

For (b), the only possibilities of eigen values are $1, \omega, \omega^2$. Now if the eigen values are only $1$ and $1$ then $A$ will satisfy $(\lambda-1)^2=0$. We already know that $A^3=I$. From these two facts it is not difficult to see that $A=kI$. Hence the only possible eigen values can be $\omega, \omega^2$. Hence (b) is true.

For(c), $A$ is definitely diagonalizable over $\mathbb C$. Is there any condition which would force a matrix to be diagonalizable over $\mathbb R$ when it is already diagonalizable over $\mathbb C$?

Answer 1

Hint for $(a)$ let $\lambda_1+\lambda_2=-1,\lambda_1\lambda_2=1\Rightarrow\lambda_1-\lambda_2=\pm\sqrt{3}i$

for $(b)$ minpoly must divide $x^3-1$, so minpoly may be $(x-1)$ or $x^2+x+1$, but as the matrix was not diagonal then minpoly is $x^2+x+1=0$.

roots of minpoly are eigen values which are $\omega,\omega^2$

so trace $\omega+\omega^2=-1$ and $\det A=\omega\times\omega^2=1$

so $b$ is correct

for $(c)$ look at $(b)$ and conclude why not diagonalizable over reals ;)

Answer 2

For (c), consider a rotation by $2\pi/3.$ This is clearly not diagonalizable over $\mathbb{R}$ (for one thing, the eigenvalues are not real).